Difference between revisions of "1994 AHSME Problems/Problem 1"

Latest revision as of 16:25, 9 January 2021

Problem

$4^4 \cdot 9^4 \cdot 4^9 \cdot 9^9=$

$\textbf{(A)}\ 13^{13} \qquad\textbf{(B)}\ 13^{36} \qquad\textbf{(C)}\ 36^{13} \qquad\textbf{(D)}\ 36^{36} \qquad\textbf{(E)}\ 1296^{26}$

Solution

Note that $a^x\times a^y=a^{x+y}$ . So $4^4\cdot 4^9=4^{13}$ and $9^4\cdot 9^9=9^{13}$ . Therefore, $4^{13}\cdot 9^{13}=(4\cdot 9)^{13}=\boxed{\textbf{(C)}\ 36^{13}}$ .

--Solution by TheMaskedMagician

1994 AHSME (Problems • Answer Key • Resources)
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Difference between revisions of "1994 AHSME Problems/Problem 1"

Latest revision as of 16:25, 9 January 2021

Problem

Solution

See Also